When a reaction is carried out at standard st | vedclass

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Question

(D) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K$.

Vedclass · Accepted Answer

$	ext{equilibrium constant } (K) = 1$

Vedclass · Answer

(D) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K$. At standard states,the reaction quotient $Q$ is equal to $1$. At equilibrium,$\Delta G = 0$. However,the question specifies standard states,implying $\Delta G^{\circ} = -RT \ln K$. For the reaction to be at equilibrium under standard conditions,$\Delta G^{\circ}$ must be $0$. Therefore,$0 = -RT \ln K$,which implies $\ln K = 0$. Thus,$K = e^0 = 1$.

When a reaction is carried out at standard states,then at equilibrium:

Similar Questions

The equilibrium constant of a reaction is related to

Assertion $(A)$: For every chemical reaction at equilibrium,standard Gibbs energy change of the reaction is zero.
Reason $(R)$: At constant temperature and pressure,chemical reactions are spontaneous in the direction of decreasing Gibbs energy.

For an equilibrium reaction,if $\Delta G^{\circ} = 0$,the equilibrium constant $K$ is equal to:

The $INCORRECT$ match in the following is

At $300 \ K$,$\Delta_{r} G^{\circ}$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $-11.5 \ kJ \ mol^{-1}$. The equilibrium constant at $300 \ K$ is approximately $(R=8.314 \ J \ mol^{-1} \ K^{-1})$.

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